Cinnamaldehyde (132.16 g/mol) is used as a flavoring agent. What mass of cinnamaldehyde must be added to 175 g of ethanol to give a solution whose boiling point is 82.57°C? (For ethanol, K_b = 1.22°C/m; boiling point of pure ethanol = 78.37°C) A) 62.4 g B) 67.8 g C) 76.2 g D) 78.5 g E) 79.6 g

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The boiling point elevation formula is $\Delta T_b = i \cdot K_b \cdot m$, where $\Delta T_b$ is the boiling point elevation, $i$ is the van't Hoff factor (which is 1 for non-electrolytes like cinnamaldehyde), $K_b$ is the ebullioscopic constant of the solvent (1.22°C/m for ethanol), and $m$ is the molality of the solution. The boiling point elevation ($\Delta T_b$) is the difference between the boiling point of the solution (82.57°C) and the boiling point of pure ethanol (78.37°C), which is 4.2°C. To find the mass of cinnamaldehyde, we first calculate the molality of the solution using the formula for boiling point elevation, then use the molality to find the moles of cinnamaldehyde, and finally convert the moles to grams.$$\Delta T_b = 4.2°C = (1) \cdot (1.22°C/m) \cdot m$$Solving for $m$, we get:$$m = \frac{4.2°C}{1.22°C/m} = 3.44 m$$Molality ($m$) is defined as moles of solute per kilogram of solvent. The mass of ethanol is 175 g, or 0.175 kg. So, the moles of cinnamaldehyde are:$$\text{moles of cinnamaldehyde} = 3.44 \cdot 0.175 = 0.602 moles$$The molar mass of cinnamaldehyde is 132.16 g/mol, so the mass of cinnamaldehyde needed is:$$\text{mass} = 0.602 \cdot 132.16 g/mol = 79.6 g$$Therefore, the correct answer is 79.6 g.